A ball is thrown into the air with an initial upward velocity of 60 ft/s. Its height (h) in feet after t seconds
is given by the function h = –16t² + 60t + 6. What will the height be at t = 3 seconds?
(1 point)
35 feet
40 feet
42 feet
45 feet
To find the height at t = 3 seconds, we substitute t = 3 into the function h = -16t^2 + 60t + 6:
h = -16(3)^2 + 60(3) + 6
h = -16(9) + 180 + 6
h = -144 + 180 + 6
h = 42
Therefore, the height at t = 3 seconds is 42 feet.
The answer is 42 feet.
Solve. x² – 121 = 0
(1 point)
0
–11
11
11, –11
To solve the equation x² - 121 = 0, we first add 121 to both sides to isolate x²:
x² - 121 + 121 = 0 + 121
x² = 121
Next, we take the square root of both sides to solve for x:
√(x²) = ±√121
x = ±11
Therefore, the solutions to the equation x² - 121 = 0 are x = 11 and x = -11.
The answer is 11, -11.